Question: The lifespans of porcupines in a particular zoo are normally distributed. The average porcupine lives $20.6$ years; the standard deviation is $1.1$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a porcupine living less than $22.8$ years.
$20.6$ $19.5$ $21.7$ $18.4$ $22.8$ $17.3$ $23.9$ $95\%$ $2.5\%$ $2.5\%$ We know the lifespans are normally distributed with an average lifespan of $20.6$ years. We know the standard deviation is $1.1$ years, so one standard deviation below the mean is $19.5$ years and one standard deviation above the mean is $21.7$ years. Two standard deviations below the mean is $18.4$ years and two standard deviations above the mean is $22.8$ years. Three standard deviations below the mean is $17.3$ years and three standard deviations above the mean is $23.9$ years. We are interested in the probability of a porcupine living less than $22.8$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $95\%$ of the porcupines will have lifespans within 2 standard deviations of the average lifespan. The remaining $5\%$ of the porcupines will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({2.5\%})$ will live less than $18.4$ years and the other half $({2.5\%})$ will live longer than $22.8$ years. The probability of a particular porcupine living less than $22.8$ years is ${95\%} + {2.5\%}$, or $97.5\%$.